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So let ABC be a right triangle at A such that BC=2AB. Find the angle $[\hat{ACB}]$

How can I find that angle without using cos, sin and other things?

Since I've already figure out how to find it using cos: here's my approach:

We denote $[\hat{ABC}]$ as $\alpha$ so $\cos\alpha=AB/BC=AB/2AB=1/2$

We do cos$^{-1}$ to find the angle ABC then we do 90-ABC to find the angle ACB.

So I'm looking for alternative way.

Thanks so

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1 Answer 1

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Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...

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I don't understand those terms, could you clarify a bit more please sir :) –  user138849 Mar 29 '14 at 13:40
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ? –  DonAntonio Mar 29 '14 at 13:42
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median –  user138849 Mar 29 '14 at 13:52
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english. –  DonAntonio Mar 29 '14 at 13:56
ookay i understand thanks anyway –  user138849 Mar 29 '14 at 13:57

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